Dr Dreidel:Is it that the simulation is actually done with replacement, and I'm thinking without?

I see it as a new puzzle the second time - 1 or 2? Or: 1, 2, 3, 4...999,997 or 999,998?

I dunno, it's not like the second box changed, either - you just know the one he took away wasn't it.

// or was it?// was he just farking with us the whole time?!?// I always knew Monty Hall was a bastard// I've been refreshing like a maniac - this is killing me// I'm like: but really, which one of us is the moran? Is it really me, again?// HAS THE WHOLE WORLD GONE CRAZY?!? AM I THE ONLY ONE WHO GIVES A SHIAT ABOUT THE PROBABILITIES?!?!?!?// slashy record

Here, maybe this awesome mspaint illustration will help:

In this example, you have chosen box one. It doesn't matter where the prize is, really - there were three boxes to start with, so there is a 33% chance your box was correct, a 66% chance it was in one of the other two boxes. So, let's draw a virtual box around the other two boxes and give it a 66% chance.

Monty then eliminates one empty, unchosen box - this, by definition, has to reside within our virtual box that has a 66% chance of holding the prize.

Having eliminated a KNOWN empty box, the one remaining box does not have its probability adjusted. There was a 66% chance of the prize residing in our virtual box before we eliminated the known bad box, and there is STILL a 66% chance of the prize residing inside of our virtual box.

I get that the Monty Hall problem is difficult for some people to understand, but I wish they could just accept that they don't understand it instead of adamantly derping about how it's 50/50. At least be smart enough to know what you don't know.

You have two boxes. One contains $1mYou choose a boxI put a 3rd, open empty box, next to the other two.I offer you a choice - keep your box or take the remaining (closed) boxDo you make a different choice than in the 1st situation? Why?

It won't matter. it's 50/50 to switch. I could make either choice and I have an equal chance of being right/wrong.

A lot of people will say you switch boxes because you'd be making a choice based on 50/50 odds, but miss that by choosing not to switch you are also making a choice based on 50/50 odds even if you originally selected it with worse odds.

Ambitwistor:DamnYankees: This guy is making the mistake of thinking the state-by-state probabilities were independent and not covariant,

His comments on that:

"Some of the bright people here at the American Center for Physics have suggested that I'm overlooking the possibility that individual states' elections can affect each other or may be mutually affected by some outside influence in a way to make them move together. That might be true, but it implies that Silver should sometimes call the entire race perfectly (as he seems to have done again this year) and at other times miss on almost every battleground state, but he would be unlikely to miss by a little. If this is the case, any one state's supposed "Chances of winning" seem to me to be meaningless, and Silver should only be able to predict how blocks of states move together. That would make his model a pretty blunt instrument, despite the fact that most of his fans act like it's the equivalent of a statistical scalpel. In fact, Silver points out in the description of his methodology that such interactions among states are accounted for in the model, which means they're included in the calculations that produce the model's output, including the chances of winning he posted for each race. So this is a pretty plausible explanation, although I don't see how I could check it easily."

So basically, he answered his own question and he doesn't even know it?